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        riddles >> putnam exam (pure math) >> transcendence of the agm
        
(Message started by: JocK on Jan 16th, 2005, 12:00pm)
Title: transcendence of the agm
Post by JocK on Jan 16th, 2005, 12:00pm
Is the arithmetic-geometric mean* of any two positive integers M and N other than M = N transcendental?



* The arithmetic-geometric mean agm(A,B) of two numbers A and B is defined as:

agm(A,B) = limn[to][infty] an

with:

a0 = A,  an+1 = (an + bn)/2
b0 = B,  bn+1 = [sqrt](an bn)
Title: Re: transcendence of the agm
Post by Barukh on Jan 18th, 2005, 10:52am
My guess is that the answer is yes, but I don't have any idea how to prove this. It looks like a very tough problem...

The only approach I tried is to show that every iteration of the AGM sequence increases the degree of the polynomial that an, bn can be. But this is not true.


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